Question 1205385
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Eggs are packed into cartons of six. A sample of 90 cartons is randomly 
selected and the number of damaged eggs in each carton counted.
<pre>
Number of damaged eggs
0
1
2
3
4
5
6

Number of Cartons
52
15
8
5
4
3
3
</pre>
Does the number of damaged eggs in a carton follow a Binomial distribution? 
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Let assume for a minute that the given distribution is a binomial with the probability of a successful outcome
(success in this problem is getting a damaged egg) &nbsp;at each individual trial &nbsp;&nbsp;p = const.



Then the probability to have 6 successful trials (six damaged eggs in a cartoon) is  &nbsp;&nbsp;{{{p^6}}} = {{{3/90}}} = {{{1/30}}},
which implies  &nbsp;&nbsp;p = {{{root(6,1/30)}}} = 0.5673  &nbsp;&nbsp;(approximately).



From the other side, &nbsp;the probability to have 0 successful trials &nbsp;(no damaged eggs in a cartoon) &nbsp;is  &nbsp;&nbsp;{{{(1-p)^6}}} = {{{52/90}}} = {{{26/45}}},
which implies  &nbsp;&nbsp;1-p = {{{root(6,26/45)}}} = 0.9126  &nbsp;&nbsp;(approximately).



But then we have this contradiction: 


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;the sum of &nbsp;p &nbsp;and (&nbsp;1-p), &nbsp;&nbsp;which is  &nbsp;&nbsp;0.5673 + 0.9126 ~ 1.4700, &nbsp;&nbsp;is &nbsp;FAR &nbsp;from to be equal &nbsp;1 &nbsp;(one).



From this reasoning, &nbsp;my conclusion is that the given distribution &nbsp;IS &nbsp;NOT &nbsp;a binomial.